I have a very basic question regarding algebraic model theory. I am trying to read Espaces de Berkovich, polytopes, squelettes et théorie des modèles (MSN) by Antoine Ducros. The relevant section is Section 0.31. Let me give the setup: Let $K$ be a valued field and let $\mathscr{C}_K$ be the category of non-trivially valued, algebraically closed field extensions of $K$. We write $\mathscr{L}_\mathrm{val}$ for the language of valued fields, having three sorts (the valued field, its residue field and the value group).

Let $\mathscr{S}$ be a product of sorts. For each model of the theory of non-trivially valued, algebraically closed field extensions of $K$, i.e. for each $F \in \mathscr{C}_K$, we can then make sense of $\mathscr{S}(F)$, and this gives a functor $\mathscr{S} \colon \mathscr{C}_K \to \mathrm{Set}$. If $\Phi(x)$ is a formula of $\mathscr{L}_\mathrm{val}$ (with parameters in $K$), whose free variables $x$ live in $\mathscr{S}$, then the subfunctor of $\mathscr{S}$ mapping $F$ to $\{x \in \mathscr{S}(F) \mid \Phi(x)\}$ is called a definable subfunctor of $\mathscr{S}$.

In general, an abstract functor from $\mathscr{C}_K$ to $\mathrm{Set}$ is called definable if it is isomorphic to a definable subfunctor of some product of sorts.

Now it is claimed that a scheme of finite type $\mathscr{X}$ over $K$ gives rise to a definable functor on $\mathscr{C}_K$. I don't understand why this is true. If $\mathscr{X}$ is affine, then I believe that the associated functor should simply be the “functor of points”. Namely, if $\mathscr{X}$ is isomorphic to $\mathrm{Spec}(K[T_1, \dots, T_n] / \langle f_1, \dots, f_r\rangle)$, then its associated functor is isomorphic to the definable subfunctor of $F \mapsto F^n$, given by the equations $f_i$. I have no idea, what is meant in the non-affine case. I would be very glad about some clarification. Also, if some of my writing above hints at some misunderstanding on my side, please let me know.


1 Answer 1


This follows by elimination of imaginaries in algebraically closed fields. Given a finite affine cover of your scheme, one obtains a functor to Set by gluing the affine charts. This is indeed not definable, but a quotient of a definable set by a definable equivalence relation. Elimination of imaginaries tells you precisely that such a quotient is in definable bijection with a definable set. You can find the details in Chapter 4 (A. Pillay, Model theory of algebraically closed fields) of the following book:

Bouscaren, Elisabeth, Introduction to model theory, Bouscaren, Elisabeth (ed.), Model theory and algebraic geometry. An introduction to E. Hrushovski’s proof of the geometric Mordell-Lang conjecture. Berlin: Springer. Lect. Notes Math. 1696, 1-18 (1998). ZBL0925.03160.

Chapters 1 and 2 also contain an introduction to general model theoretic concepts including elimination of imaginaries (Chapter 2).

  • 1
    $\begingroup$ I edited in links to the book and the individual chapters. Since you later cite Chapter 1 separately, I think you didn't mean your reference and corresponding ZBL link to refer only to the first chapter, but rather to the whole book. That's ZBL0920.03046. $\endgroup$
    – LSpice
    Jun 6, 2020 at 18:05
  • $\begingroup$ Thank you very much. $\endgroup$
    – Cubikova
    Jun 6, 2020 at 18:18
  • $\begingroup$ If I understand “gluing affine charts” correctly, then you are saying that also for general $\mathscr{X}$, the associated functor is the functor of points, sending $F$ to $\mathscr{X}(F) = \mathrm{Hom}_K(\mathrm{Spec}(F), \mathscr{X})$, right? Could you maybe indicate for some simple non-affine scheme like $\mathbf{P}^1_K$, how the bijection that you get to a definable functor would look like? $\endgroup$ Jun 6, 2020 at 18:22
  • $\begingroup$ By the way, thank you for the hints to the literature, I will definitely have a look at it. $\endgroup$ Jun 6, 2020 at 18:24
  • $\begingroup$ Indeed, the functor is isomorphic to the functor of points. In the case of $\mathbf{P}_K^n$, the functor of points is isomorphic to the functor sending $F$ to the set of $n$-dimensional subspaces of $F^{n+1}$. The usual equivalence relation on $F^{n+1}$ given by $(x_0,\ldots, x_n)\sim(y_0,\ldots,y_n)$ if and only if there is $\lambda\in F^*$ such that $\lambda x_i=y_i$ for each $i\in \{0,\ldots,n\}$ is the corresponding definable equivalence relation. $\endgroup$
    – Cubikova
    Jun 8, 2020 at 9:47

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